Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose $X_1,\ldots,X_n$ are a random sample of $N(0,\sigma^2)$. How can find $$ E\left(\frac{\bigl(\sum_{i=1}^nX_i\bigr)^2}{\sum_{i=1}^n X_i^2}\right) $$

share|cite|improve this question
Expand the numerator and the first term will be very nice. The second term can be written as a sum of products of zero-mean uncorrelated (but not independent!) random variables (why?). Use a standard fact about such products to conclude. – cardinal Sep 11 '12 at 2:18
We can also use Basu theorem for solve this where denominator is complete statistics and numerator is anciliary statistics – Amirhossein Jan 11 '14 at 22:48

2 Answers 2

up vote 4 down vote accepted

Assuming $X_i$ are independent identically distributed normals with zero mean.

Using $\frac{1}{\lambda} = \int_0^\infty \mathrm{e}^{-\lambda t} \mathrm{d} t$: $$\begin{eqnarray} \mathbb{E}\left( \frac{\left(\sum_{i=1}^n X_i\right)^2}{\sum_{i=1}^n X_i^2} \right) &=& \int_0^\infty \mathbb{E}\left( \sum_{i=1}^n \sum_{j=1}^n X_i X_j \mathrm{e}^{-t \sum_{k=1}^n X_k^2} \right) \mathrm{d} t \\ &\stackrel{\text{symmetry}}{=}& \int_0^\infty \sum_{i=1}^n \mathbb{E}\left( X_i^2 \mathrm{e}^{-t \sum_{k=1}^n X_k^2}\right) \mathrm{d} t \\ &\stackrel{\text{indep.}}{=}& \int_0^\infty \sum_{i=1}^n \mathbb{E}\left(X_i^2 \mathrm{e}^{-t X_i^2} \right) \prod_{k=1,k\not= i}^n \mathbb{E}\left(\mathrm{e}^{-t X_k^2} \right) \mathrm{d} t \\ &=& \int_0^\infty \left(-\frac{\mathrm{d}}{\mathrm{d}t} \left(\mathbb{E}\left(\mathrm{e}^{-t X^2} \right)\right)^n \right)\mathrm{d} t \\&=& \left.-\left(\mathbb{E}\left(\mathrm{e}^{-t X^2} \right)\right)^n\right|_{t \to 0^+}^{t \to \infty} = 1 \end{eqnarray} $$

share|cite|improve this answer

Expand the numerator. The expectation value for the mixed terms vanishes by symmetry. The sum of the remaining terms is the denominator, so the remaining fraction is $1$, and so is its expectation value.

share|cite|improve this answer
+1 Nice! Simpler is better. – Sasha Sep 11 '12 at 3:04
I see no way to make this answer better than it is. – Did Sep 11 '12 at 16:31
(+1) Good answer. – cardinal Sep 11 '12 at 16:45

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.